std::numeric_limits::epsilon

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static T epsilon()
(C++11 前)
static constexpr T epsilon()
(C++11 起)
返回的机器ε之间的差异由浮点类型1.0T和下一个值表示的,即,。这是唯一有意义的,如果std::numeric_limits<T>::is_integer == false.
原文:
Returns the machine epsilon, that is, the difference between 1.0 and the next value representable by the floating-point type T. It is only meaningful if std::numeric_limits<T>::is_integer == false.
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目录

[编辑] 返回值

T std::numeric_limits<T>::epsilon()
/* non-specialized */ T();
bool false
char 0
signed char 0
unsigned char 0
wchar_t 0
char16_t 0
char32_t 0
short 0
unsigned short 0
int 0
unsigned int 0
long 0
unsigned long 0
long long 0
unsigned long long 0
float FLT_EPSILON
double DBL_EPSILON
long double LDBL_EPSILON

[编辑] 例外

noexcept指定:  
noexcept
  (C++11 起)

[编辑] 示例

演示简单地使用机器精度浮点值比较.
原文:
Demonstrates the simplistic use of machine epsilon to compare floating-point values.
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#include <cmath>
#include <limits>
#include <iomanip>
#include <iostream>
#include <type_traits>
 
template<class T>
typename std::enable_if<!std::numeric_limits<T>::is_integer, bool>::type
    almost_equal(T x, T y, int ulp)
{
    // the machine epsilon has to be scaled to the magnitude of the larger value
    // and multiplied by the desired precision in ULPs (units in the last place)
    return std::abs(x-y) <=   std::numeric_limits<T>::epsilon()
                            * std::max(std::abs(x), std::abs(y))
                            * ulp;
}
int main()
{
    double d1 = 0.2;
    double d2 = 1 / std::sqrt(5) / std::sqrt(5);
 
    if(d1 == d2)
            std::cout << "d1 == d2\n";
    else
            std::cout << "d1 != d2\n";
 
    if(almost_equal(d1, d2, 2))
            std::cout << "d1 almost equals d2\n";
    else
            std::cout << "d1 does not almost equal d2\n";
}

输出:

d1 != d2
d1 almost equals d2

[编辑] 另请参阅

(C++11)
(C++11)
给定值的下一个可表示浮点值
原文:
next representable floating point value towards the given value
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(函数) [edit]